Question

11-18: Find the differential of the function

16. \(y = \sqrt {1 + \cos \theta } \)

Short answer

The differential of the function is \(dy = - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}d\theta \).

Step-by-step solution

Definition of differentials

The equation establishes the differential \(dy\)with respect to \(dx\) as shown below:

\(dy = f'\left( x \right)dx\)

Where \(dx\) represent the independent variableand \(dy\) represent the dependent variable.

Determine the differential of the function

The given function is \(y = f\left( \theta \right) = \sqrt {1 + \cos \theta } \).

Evaluate the derivative of the function as shown below:

\(\begin{aligned}f'\left( \theta \right) &= \frac{d}{{d\theta }}\left( {\sqrt {1 + \cos \theta } } \right)\\ &= \frac{d}{{d\theta }}{\left( {1 + \cos \theta } \right)^{\frac{1}{2}}}\\ &= \frac{1}{2}{\left( {1 + \cos \theta } \right)^{ - \frac{1}{2}}}\left( { - \sin \theta } \right)\\ &= - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}\end{aligned}\)

Obtain the differential of the function as shown below:

\(\begin{aligned}dy &= f'\left( \theta \right)d\theta \\dy &= - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}d\theta \end{aligned}\)

Thus, the differential of the function is \(dy = - \frac{{\sin \theta }}{{2\sqrt {1 + \cos \theta } }}d\theta \).